Differential EquationsmediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Circle Radius ≤ 6, Integer p Count = 11 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let y=y(x)y=y(x) be the solution of xsin⁡(yx)dy=(ysin⁡(yx)−x)dxx\sin\left(\dfrac{y}{x}\right)dy=\left(y\sin\left(\dfrac{y}{x}\right)-x\right)dx, y(1)=π2y(1)=\dfrac{\pi}{2}, and let α=cos⁡(y(e12)e12)\alpha=\cos\left(\dfrac{y(e^{12})}{e^{12}}\right). Then the number of integral values of pp for which x2+y2−2px+2py+α+2=0x^2+y^2-2px+2py+\alpha+2=0 represents a circle of radius r≤6r\le6 is
Solution
Answer: 6 (± 0.01)
Step 1: Homogeneous. Let y=vx⇒dy=v dx+x dvy=vx\Rightarrow dy=v\,dx+x\,dv. Then xsin⁡v (v dx+x dv)=(vxsin⁡v−x)dxx\sin v\,(v\,dx+x\,dv)=(vx\sin v-x)dx. Step 2: Divide by xx: sin⁡v (v dx+x dv)=(vsin⁡v−1)dx⇒vsin⁡v dx+xsin⁡v dv=vsin⁡v dx−dx⇒xsin⁡v dv=−dx\sin v\,(v\,dx+x\,dv)=(v\sin v-1)dx\Rightarrow v\sin v\,dx+x\sin v\,dv=v\sin v\,dx-dx\Rightarrow x\sin v\,dv=-dx. Step 3: sin⁡v dv=−dxx⇒∫sin⁡v dv=−∫dxx⇒−cos⁡v=−ln⁡∣x∣+C⇒cos⁡ ⁣(yx)=ln⁡∣x∣+C1\sin v\,dv=-\dfrac{dx}{x}\Rightarrow\displaystyle\int\sin v\,dv=-\int\dfrac{dx}{x}\Rightarrow-\cos v=-\ln|x|+C\Rightarrow\cos\!\left(\dfrac{y}{x}\right)=\ln|x|+C_1. Step 4: y(1)=π2y(1)=\dfrac{\pi}{2}: cos⁡π2=ln⁡1+C1⇒0=0+C1⇒C1=0\cos\dfrac{\pi}{2}=\ln 1+C_1\Rightarrow 0=0+C_1\Rightarrow C_1=0, ∴cos⁡ ⁣(yx)=ln⁡x\therefore\cos\!\left(\dfrac{y}{x}\right)=\ln x. Step 5: At x=e1/2x=e^{1/2}: cos⁡ ⁣(yx)=ln⁡e1/2=12\cos\!\left(\dfrac{y}{x}\right)=\ln e^{1/2}=\dfrac12. By definition α=cos⁡ ⁣(y(e1/2)e1/2)=12\alpha=\cos\!\left(\dfrac{y(e^{1/2})}{e^{1/2}}\right)=\dfrac12. Step 6: x2+y2−2px+2py+(12+2)=0x^2+y^2-2px+2py+\left(\dfrac12+2\right)=0, centre (p,−p)(p,-p), r2=p2+p2−52=2p2−52r^2=p^2+p^2-\dfrac52=2p^2-\dfrac52. Step 7: Circle needs r2>0r^2>0 and r≤6⇒0<2p2−52≤36⇒54<p2≤774⇒1.118<∣p∣≤4.39r\le6\Rightarrow 0<2p^2-\dfrac52\le36\Rightarrow \dfrac54<p^2\le\dfrac{77}{4}\Rightarrow 1.118<|p|\le4.39. Step 8: Integer pp: ∣p∣∈{2,3,4}⇒p=±2,±3,±4⇒6|p|\in\{2,3,4\}\Rightarrow p=\pm2,\pm3,\pm4\Rightarrow 6 values. Correct answer: 6
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